Editing Platonic solid (section)

== Classification ==
The classical result is that only five convex regular polyhedra exist. Two common arguments below demonstrate no more than five Platonic solids can exist, but positively demonstrating the existence of any given solid is a separate question—one that requires an explicit construction.

=== Geometric proof ===
{| class="wikitable floatright" style="text-align:center"
|+ Polygon nets around a vertex
|- style="vertical-align:bottom;"
| [[File:Polyiamond-3-1.svg|80px]]<BR/>{3,3}<BR/>Defect 180°
| [[File:Polyiamond-4-1.svg|80px]]<BR/>{3,4}<BR/>Defect 120°
| [[File:Polyiamond-5-4.svg|80px]]<BR/>{3,5}<BR/>Defect 60°
| style="background-color:#e0e0ff;" | [[File:Polyiamond-6-11.svg|80px]]<BR/>{3,6}<BR/>Defect 0°
|- style="vertical-align:bottom;"
| [[File:TrominoV.svg|80px]]<BR/>{4,3}<BR/>Defect 90°
| style="background-color:#e0e0ff;" | [[File:Square tiling vertfig.svg|80px]]<BR/>{4,4}<BR/>Defect 0°
| [[File:Pentagon_net.svg|80px]]<BR/>{5,3}<BR/>Defect 36°
| style="background-color:#e0e0ff;" | [[File:Hexagonal tiling vertfig.svg|80px]]<BR/>{6,3}<BR/>Defect 0°
|-
| colspan=4 | A vertex needs at least 3 faces, and an [[angle defect]]. <BR/>A 0° angle defect will fill the Euclidean plane with a regular tiling. <BR/>By [[angular defect#Descartes' theorem|Descartes' theorem]], the number of vertices is 720°/''defect''.
|}

The following geometric argument is very similar to the one given by [[Euclid]] in the [[Euclid's Elements|''Elements'']]:

{{ordered list
| Each vertex of the solid must be a vertex for at least three faces.
| At each vertex of the solid, the total, among the adjacent faces, of the angles between their respective adjacent sides must be strictly less than 360°. The amount less than 360° is called an [[angle defect]].
| The angles at all vertices of all faces of a Platonic solid are identical: each vertex of each face must contribute less than {{sfrac|360°|3}}&nbsp;=&nbsp;120°.
| Regular polygons of [[Hexagon|six]] or more sides have only angles of 120° or more, so the common face must be the triangle, square, or pentagon. For these different shapes of faces the following holds:
; [[Triangle|Triangular]] faces: Each vertex of a regular triangle is 60°, so a shape may have three, four, or five triangles meeting at a vertex; these are the tetrahedron, octahedron, and icosahedron respectively.
; [[Square (geometry)|Square]] faces: Each vertex of a square is 90°, so there is only one arrangement possible with three faces at a vertex, the cube.
; [[Pentagon]]al faces: Each vertex is 108°; again, only one arrangement of three faces at a vertex is possible, the dodecahedron.

Altogether this makes five possible Platonic solids.
}}

=== Topological proof ===
A purely [[topology|topological]] proof can be made using only combinatorial information about the solids. The key is [[Euler characteristic|Euler's observation]] that ''V''&nbsp;−&nbsp;''E''&nbsp;+&nbsp;''F''&nbsp;=&nbsp;2, and the fact that ''pF''&nbsp;=&nbsp;2''E''&nbsp;=&nbsp;''qV'', where ''p'' stands for the number of edges of each face and ''q'' for the number of edges meeting at each vertex. Combining these equations one obtains the equation

{{Hamiltonian_platonic_graphs.svg}}

<math display="block">\frac{2E}{q} - E + \frac{2E}{p} = 2.</math>

Simple algebraic manipulation then gives

<math display="block">{1 \over q} + {1 \over p}= {1 \over 2} + {1 \over E}.</math>

Since ''E'' is strictly positive we must have

<math display="block">\frac{1}{q} + \frac{1}{p} > \frac{1}{2}.</math>

Using the fact that ''p'' and ''q'' must both be at least 3, one can easily see that there are only five possibilities for {''p'',&nbsp;''q''}:
{{block indent|{3,&nbsp;3}, {4,&nbsp;3}, {3,&nbsp;4}, {5,&nbsp;3}, {3,&nbsp;5}.}}